
Let \({\Lambda _k} \triangleq \frac{{{P_k}D_{{\rm {u}}k}^{ - {\alpha _k}}\left( {{m_{sk}} - 1} \right){{\bar{z}}_k}}}{{{m_{fk}}}}\) . With the help of [1]}, the integration part in \(I_1\)  can be solved. Thus, we can re-write \(I_1\)  as
\(&{I_1} = \frac{1}{{2\pi i}}{\left( {\frac{{{P_k}D_{{\rm {u}}k}^{ - {\alpha _k}}}}{{{m_{fk}}\gamma }}} \right)^{{m_{fk}} + {m_{sk}}}}\int _{\cal L} {\Gamma \!\left( { - {s_1}} \right)}\\&\times \!{\left( {\frac{1}{{{P_{Ik}}{\eta _k}}}} \right)^{{s_1}}}{\left( {\frac{{{P_k}D_{{\rm {u}}k}^{ - {\alpha _k}}\left( {{m_{sk}} - 1} \right){{\bar{z}}_k}}}{{{m_{fk}}\gamma }} + {\sigma ^2}} \right)^{ - {m_{fk}} - {m_{sk}}}}\\&\times \!{\sigma ^{2{s_1} + 2{N_{Ik}} + 2{m_{fk}}}}B\!\left( {{s_1}\! + \!{N_{Ik}},{m_{sk}} - {s_1} - {N_{Ik}}} \right)\\&\times \!F\!\left(\! {{m_{fk}}\! +\! {m_{sk}},{s_1} \!+\! {N_{Ik}};{m_{sk}};1 \!-\! \frac{{{\sigma ^2}}}{{\frac{{{\Lambda _k}}}{\gamma } + {\sigma ^2}}}} \!\right){\rm {d}}{s_1}.\) 
